Hopf subalgebras of the Hopf algebra of rooted trees coming from

Dyson-Schwinger equations and Lie algebras of Fa di Bruno type

Loıc Foissy

Contents

1 The Hopf algebra of rooted trees and Dyson-Schwinger equations 21.1 The Connes-Kreimer Hopf algebra . . . . . . . . . . . . . . . . . . . . . . . . . . 21.2 Gradation of H and completion . . . . . . . . . . . . . . . . . . . . . . . . . . . . 41.3 1-cocycle of H and Dyson-Schwinger equations . . . . . . . . . . . . . . . . . . . 5

2 Formal series giving a Hopf subalgebras 72.1 Statement of the main theorem . . . . . . . . . . . . . . . . . . . . . . . . . . . . 72.2 Proof of 1 =⇒ 2 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 72.3 Proof of 2 =⇒ 1 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 82.4 What is Hα,β ? . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 11

3 FdB Lie algebras 123.1 Definitions and first properties . . . . . . . . . . . . . . . . . . . . . . . . . . . . 123.2 Case where µ3 = 1 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 153.3 Case where µ3 = 0 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 17

4 Dual of enveloping algebras of FdB Lie algebras 184.1 The corolla Lie algebra . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 184.2 The third FdB Lie algebra . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 19

Introduction

The Connes-Kreimer algebra H of rooted trees is introduced in [7]. This graded Hopf algebrais commutative, non cocommutative, and is given a linear basis by the set of rooted forests. Aparticularly important operator of H is the grafting on a root B+, which satisfies the followingequation:

∆ ◦B+(x) = B+(x)⊗ 1 + (Id⊗B+) ◦∆(x).

In other terms, B+ is a 1-cocycle for the Cartier-Quillen cohomology of coalgebras. Moreover,the couple (H, B+) satisfies a universal property, see theorem 3 of the present text.

We consider here a family of subalgebras of H, associated to the combinatorial Dyson-Schwinger equation [1, 8, 9]:

X = B+(f(X)),

where f =∑

pnhn is a formal series such that p0 = 1, and X is an element of the completionof H for the topology given by the gradation of H. This equation admits a unique solutionX =

∑xn, where xn is, for all n ≥ 1, a linear span of rooted trees of weight n, inductively given

by: x1 = p0 q ,

xn+1 =n∑

k=1

∑a1+...+ak=n

pkB+(xa1 . . . xak

).

1

We denote by Hf the subalgebra of H generated by the xn’s.For the usual Dyson-Schwinger equation, f = (1−h)−1. It turns out that, in this case, Hf is

a Hopf subalgebra. This is not the case in general; we characterise here the formal series f suchthat Hf is Hopf. Namely, Hf is a Hopf subalgebra of H if, and only if, there exists (α, β) ∈ K2,

such that f(h) = 1 if α = 0, or f(h) = eαh if β = 0, or f(h) = (1 − αβh)−1β if αβ 6= 0. We

obtain in this way a two-parameters family Hα,β of Hopf subalgebras of H and we explicitelydescribe a system of generator of these algebras. In particular, if α = 0, then Hα,β = K[ q ]; ifα 6= 0, then Hα,β = H1,β .

The Hopf algebraHα,β is commutative, graded and connected. By the Milnor-Moore theorem[10], its dual is the enveloping algebra of a Lie algebra gα,β. Computing this Lie algebra, we findthree isomorphism classes of Hα,β’s:

1. H0,1, equal to K[ q ].2. H1,−1, the subalgebra of ladders, isomorphic to the Hopf algebra of symmetric functions.

3. The H1,β ’s, with β 6= −1, isomorphic to the Faa di Bruno Hopf algebra.

Note that non commutative versions of these results are exposed in [5].

In particular, if Hα,β is non cocommutative, it is isomorphic to the Faa di Bruno Hopfalgebra. We try to explain this fact in the third section of this text. The dual Lie algebra gα,β

satisfies the following properties:

1. gα,β is graded and connected.

2. The homogeneous component g(n) of degree n of g is 1-dimensional for all n ≥ 1.

Moreover, if Hα,β is not cocommutative, then [g(1), g(n)] 6= (0) if n ≥ 2. Such a Lie algebra willbe called a FdB Lie algebra. We prove here that there exists, up to an isomorphism, only threeFdB Lie algebras:

1. The Faa di Bruno Lie algebra, which is the Lie algebra of the group of formal diffeomor-phisms tangent to the identity at 0.

2. The Lie algebra of corollas.

3. A third Lie algebra.

In particular, with a stronger condition of non commutativity, a FdB Lie algebra is isomorphicto the Faa di Bruno Lie algebra, and this result can be applied to all H1,β ’s when β 6= −1. Thedual of the enveloping algebras of the two others FdB Lie algebras can also be embedded in H,using corollas for the second, giving in a certain way a limit of H1,β when β goes to ∞, and thethird one with a different construction.

Notation. We denote by K a commutative field of characteristic zero.

1 The Hopf algebra of rooted trees and Dyson-Schwinger equa-tions

1.1 The Connes-Kreimer Hopf algebra

Definition 1 [12, 13]

1. A rooted tree is a finite graph, connected and without loops, with a special vertex calledthe root.

2

2. The weight of a rooted tree is the number of its vertices.

3. The set of rooted trees will be denoted by T .

Examples. Rooted trees of weight ≤ 5:

q , qq , q∨qq, qqq , q∨qq q

, q∨qqq,

q∨qq q , qqqq , q∨qq�Hq q

, q∨qq qq, q∨qq qq

, q∨qq∨qq, q∨qqqq

,q∨qq qq ,

q∨qq qq , qqq∨q q, qqqqq .

The Connes-Kreimer Hopf algebra of rooted trees H is introduced in [2]. As an algebra, His the free associative, commutative, unitary algebra generated by the elements of T . In otherterms, a K-basis of H is given by rooted forests, that is to say non necessarily connected graphsF such that each connected component of F is a rooted tree. The set of rooted forests will bedenoted by F . The product of H is given by the concatenation of rooted forests, and the unitis the empty forest, denoted by 1.

Examples. Rooted forests of weight ≤ 4:

1, q , q q , qq , q q q , qq q , q∨qq, qqq , q q q , qq q q , qq qq , q∨qq q , qqq q , q∨qq q

, q∨qqq,

q∨qq q , qqqq .

In order to make H a bialgebra, we now introduce the notion of cut of a tree t. A non totalcut c of a tree t is a choice of edges of t. Deleting the chosen edges, the cut makes t into a forestdenoted by W c(t). The cut c is admissible if any oriented path1 in the tree meets at most onecut edge. For such a cut, the tree of W c(t) which contains the root of t is denoted by Rc(t)and the product of the other trees of W c(t) is denoted by P c(t). We also add the total cut,which is by convention an admissible cut such that Rc(t) = 1 and P c(t) = W c(t) = t. The setof admissible cuts of t is denoted by Adm∗(t). Note that the empty cut of t is admissible; wedenote Adm(t) = Adm∗(t)− {empty cut, total cut}.

Example. Let us consider the rooted tree t = q∨qqq. As it as 3 edges, it has 23 non total cuts.

cut c q∨qqq q∨qqq q∨qqq q∨qqq q∨qqq q∨qqq q∨qqq q∨qqqtotal

Admissible? yes yes yes yes no yes yes no yes

W c(t) q∨qqq qq qq q q∨qq qqq q q q qq qq q q qq q q q q q q q∨qqqRc(t) q∨qqq qq q∨qq qqq × q qq × 1

P c(t) 1 qq q q × qq q q q × q∨qqqThe coproduct of H is defined as the unique algebra morphism from H to H⊗H such that, forall rooted tree t ∈ T :

∆(t) =∑

c∈Adm∗(t)

P c(t)⊗Rc(t) = t⊗ 1 + 1⊗ t +∑

c∈Adm(t)

P c(t)⊗Rc(t).

As H is the polynomial algebra generated by T , this makes sense.

Example.

∆( q∨qqq) = q∨qqq

⊗ 1 + 1⊗ q∨qqq+ qq ⊗ qq + q ⊗ q∨qq

+ q ⊗ qqq + qq q ⊗ q + q q ⊗ qq .1The edges of the tree are oriented from the root to the leaves.

3

Theorem 2 [2] With this coproduct, H is a bialgebra. The counit of H is given by:

ε :{

H −→ KF ∈ F −→ δ1,F .

The antipode is the algebra endomorphism defined for all t ∈ T by:

S(t) = −∑

c non total cut of t

(−1)ncW c(t),

where nc is the number of cut edges in c.

1.2 Gradation of H and completion

We graduate H by putting the forests of weight n homogeneous of degree n. We denote by H(n)the homogeneous component of H of degree n. Then H is a graded bialgebra, that is to say:

1. For all i, j ∈ N, H(i)H(j) ⊆ H(i + j).

2. For all k ∈ N, ∆(H(k)) ⊆∑

i+j=k

H(i)⊗H(j).

We define, for all x, y ∈ H:val(x) = max

n ∈ N / x ∈⊕k≥n

H(k)

,

d(x, y) = 2−val(x−y),

with the convention 2−∞ = 0. Then d is a distance on H. The metric space (H, d) is notcomplete: its completion will be denoted by H. As a vector space:

H =∏n∈N

H(n).

The elements of H will be denoted∑

xn, where xn ∈ H(n) for all n ∈ N. The productm : H ⊗ H −→ H is homogeneous of degree 0, so is continuous. So it can be extended fromH ⊗ H to H, which is then an associative, commutative algebra. Similarly, the coproduct of Hcan be extended in an application:

∆ : H −→ H⊗H =∏

i,j∈NH(i)⊗H(j).

Let f(h) =∑

pnhn ∈ K[[h]] be any formal series, and let X =∑

xn ∈ H, such that x0 = 0.The series of H of term pnXn is Cauchy, so converges. Its limit will be denoted by f(X). Inother terms, f(X) =

∑yn, with:

yn =n∑

k=1

∑a1+...+ak=n

pkxa1 . . . xak.

Remark. If f(h) ∈ K[[h]], g(h) ∈ K[[h]], without constant terms, and X ∈ H, withoutconstant terms, it is easy to show that (f ◦ g)(X) = f(g(X)).

4

1.3 1-cocycle of H and Dyson-Schwinger equations

We define the operator B+ : H −→ H, sending a forest t1 . . . tn to the tree obtained by grafting

t1, . . . , tn to a common root. For example, B+( qq q) = q∨qqq. This operator satisfies the following

relation: for all x ∈ H,

∆ ◦B+(x) = B+(x)⊗ 1 + (Id⊗B+) ◦∆(x). (1)

This means that B+ is a 1-cocycle for a certain cohomology, namely the Cartier-Quillen coho-mology for coalgebras, dual notion of the Hochschild cohomology [2]. Moreover, (H, B+) satisfiesthe following universal property:

Theorem 3 (Universal property) Let A be a commutative algebra and let L : A −→ Abe a linear application.

1. There exists a unique algebra morphism φ : H −→ A, such that φ ◦B+ = L ◦ φ.

2. If moreover A is a Hopf algebra and L satisfies (1), then φ is a Hopf algebra morphism.

The operator B+ is homogeneous of degree 1, so is continuous. As a consequence, it can beextended as an operator B+ : H −→ H. This operator still satisfies (1).

Definition 4 [1, 8, 9] Let f ∈ K[[h]]. The Dyson-Schwinger equation associated to f is:

X = B+(f(X)), (2)

where X is an element of H, without constant term.

Proposition 5 The Dyson-Schwinger equation associated to f(h) =∑

pnhn admits a uniquesolution X =

∑xn, inductively defined by:

x0 = 0,x1 = p0 q ,

xn+1 =n∑

k=1

∑a1+...+ak=n

pkB+(xa1 . . . xak

).

Proof. It is enough to identify the homogeneous components of the two members of (2). �

Definition 6 The subalgebra of H generated by the homogeneous components xn’s of theunique solution X of the Dyson-Schwinger equation (2) associated to f will be denoted by Hf .

The aim of this text is to give a necessary and suffisant condition on f for Hf to be a Hopfsubalgebra of H.

Remarks.

1. If f(0) = 0, the unique solution of (2) is 0. As a consequence, Hf = K is a Hopf subalgebra.

2. For all α ∈ K, if X =∑

xn is the solution of the Dyson-Schwinger equation associated tof , the unique solution of the Dyson-Schwinger equation associated to αf is

∑αnxn. As

a consequence, if α 6= 0, Hf = Hαf . We shall then suppose in the sequel that p0 = 1. Inthis case, x1 = q .

Examples.

5

1. We take f(h) = 1 + h. Then x1 = q , x2 = qq , x3 = qqq , x4 = qqqq . More generally, xn is theladder with n vertices, that is to say (B+)n(1) (definition 7). As a consequence, for alln ≥ 1:

∆(xn) =∑

i+j=n

xi ⊗ xj .

So H1+h is Hopf. Moreover, it is cocommutative.

2. We take f(h) = 1 + h + h2 + 2h3. Then:x1 = q ,x2 = qq ,x3 = q∨qq

+ qqq ,x4 = 2 q∨qq q

+ 2 q∨qqq+ q∨qq q + qqqq .

Hence:

∆(x1) = x1 ⊗ 1 + 1⊗ x1,

∆(x2) = x2 ⊗ 1 + 1⊗ x2 + x1 ⊗ x1,

∆(x3) = x3 ⊗ 1 + 1⊗ x3 + x21 ⊗ x1 + 3x1 ⊗ x2 + x2 ⊗ x1,

∆(x4) = x4 ⊗ 1 + 1⊗ x4 + 10x21 ⊗ x2 + x3

1 ⊗ x1 + 3x2 ⊗ x2

+2x1x2 ⊗ x1 + x3 ⊗ x1 + x1 ⊗ (8 q∨qq+ 5 qqq),

so Hf is not Hopf.

We shall need later these two families of rooted trees:

Definition 7 Let n ≥ 1.

1. The ladder ln of weight n is the rooted tree (B+)n(1). For example:

l1 = q , l2 = qq , l3 = qqq , l4 = qqqq .

2. The corolla cn of weight n is the rooted tree B+( qn−1). For example:

c1 = q , c2 = qq , c3 = q∨qq, c4 = q∨qq q

.

The following lemma is an immediate corollary of proposition 5:

Lemma 8 The coefficient of the ladder of weight n in xn is pn−11 . The coefficient of the

corolla of weight n in xn is pn−1.

Using (1):

Lemma 9 For all n ≥ 1:

1. ∆(ln) =n∑

i=0

li ⊗ ln−i, with the convention l0 = 1.

2. ∆(cn) = cn ⊗ 1 +n−1∑i=0

(n− 1

i

) q i ⊗ cn−i.

6

2 Formal series giving a Hopf subalgebras

2.1 Statement of the main theorem

Theorem 10 Let f(h) ∈ K[[h]], such that f(0) = 1. The following assertions are equivalent:

1. Hf is a Hopf subalgebra of H.

2. There exists (α, β) ∈ K2, such that (1− αβh)f ′(h) = αf(h).

3. There exists (α, β) ∈ K2, such that f(h) = 1 if α = 0, or f(h) = eαh if β = 0, orf(h) = (1− αβh)−

1β if αβ 6= 0.

Clearly, the second and the third points are equivalent.

2.2 Proof of 1 =⇒ 2

We suppose that Hf is Hopf.

Lemma 11 Let us suppose that p1 = 0. Then f(h) = 1, so 2 holds with α = 0.

Proof. Let us suppose that pn 6= 0 for a certain n ≥ 2. Let us choose a minimal n. Thenx1 = q , x2 = . . . = xn = 0, and xn+1 = pncn+1. So:

∆(xn+1) = xn+1 ⊗ 1 + 1⊗ xn+1 +n∑

i=1

(n

i

)pn q i ⊗ cn+1−i ∈ Hf ⊗Hf .

In particular, for i = n− 1, c2 = qq ∈ Hf , so x2 6= 0: contradiction. �

We now assume that p1 6= 0. Let Z q : H −→ K, defined by Z q (F ) = δ q ,F for all F ∈ F .This application Z q is homogeneous of degree −1, so is continuous and can be extended in anapplication Z q : H −→ K. We put (Z q ⊗ Id) ◦∆(X) =

∑yn, where X is the unique solution of

(2). A direct computation shows that yn can be computed by induction with:

y0 = 1,

yn+1 =n∑

k=1

∑a1+...+ak=n

(k + 1)pk+1B+(xa1 . . . xak

)

+n∑

k=1

∑a1+...+ak=n

kpkB+(ya1xa2 . . . xak

).

As Hf is Hopf, yn ∈ Hf for all n ∈ N. Moreover, yn is a linear span of rooted trees of weight n,so is a multiple of xn: we put yn = αnxn.

Let us consider the coefficient of the ladder of weight n in yn. By lemma 8, this is αnpn−11 .

So, for all n ≥ 1:pn1αn+1 = 2pn−1

1 p2 + pn1αn.

As α1 = p1, for all n ≥ 1, αn = p1 + 2p2

p1(n− 1). Let us consider the coefficient of the corolla of

weight n in yn. By lemma 8, this is αnpn. So, for all n ≥ 1:

αnpn = (n + 1)pn+1 + npnp1.

Summing all these relations, putting α = p1 and β = 2p2

p1− 1, we obtain (1−αβh)f ′(h) = f(h),

so (2) holds.

7

2.3 Proof of 2 =⇒ 1

Let us suppose 2 or, equivalently, 3. We now writeHα,β instead ofHf . We first give a descriptionof the xn’s.

Definition 12

1. Let F ∈ F . The coefficient sF is inductively computed by:s q = 1,

sta11 ...t

akk

= a1! . . . ak!sa1t1

. . . saktk

,

sB+(ta11 ...t

akk ) = a1! . . . ak!sa1

t1. . . sak

tk,

where t1, . . . , tk are distinct elements of T .

2. Let F ∈ F . The coefficient eF is inductively computed by:

e q = 1,

eta11 ...t

akk

=(a1 + . . . + ak)!

a1! . . . ak!ea1t1

. . . eaktk

,

eB+(ta11 ...t

akk ) =

(a1 + . . . + ak)!a1! . . . ak!

ea1t1

. . . eaktk

,

where t1, . . . , tk are distinct elements of T .

Remarks.

1. The coefficient sF is the number of symmetries of F , that is to say the number of graphautomorphisms of F respecting the roots.

2. The coefficient eF is the number of embeddings of F in the plane, that is to say the numberof planar forests which underlying rooted forest is F .

We now give β-equivalents of these coefficients. For all k ∈ N∗, we put [k]β = 1 + β(k − 1)and [k]β ! = [1]β . . . [k]β. We then inductively define [sF ]β and [eF ]β for all F ∈ F by:

[s q ]β = 1,

[sta11 ...t

akk

]β = [a1]β ! . . . [ak]β ![st1 ]a1β . . . [stk ]ak

β ,

[sB+(ta11 ...t

akk )]β = [a1]β ! . . . [ak]β ![st1 ]

a1β . . . [stk ]ak

β ,[e q ] = 1,

[eta11 ...t

akk

]β =[a1 + . . . + ak]β![a1]β! . . . [ak]β!

[et1 ]a1β . . . [etk ]ak

β ,

[eB+(ta11 ...t

akk )]β =

[a1 + . . . + ak]β![a1]β! . . . [ak]β!

[et1 ]a1β . . . [etk ]ak

β ,

where t1, . . . , tk are distinct elements of T . In particular, [st]1 = st and [et]1 = et, whereras[st]0 = 1 and [et]0 = 1 all t ∈ T .

8

Examples.t st [st]β et [et]βq 1 1 1 1qq 1 1 1 1q∨qq

2 (1 + β) 1 1qqq 1 1 1 1q∨qq q6 (1 + β)(1 + 2β) 1 1

q∨qqq1 1 2 (1 + β)q∨qq q 2 (1 + β) 1 1

qqqq 1 1 1 1

Proposition 13 For all n ∈ N∗, in Hα,β, xn = αn−1∑

t∈T , weight(t)=n

[st]β[et]βst

t.

Examples.

x1 = q ,x2 = α qq ,x3 = α2

((1 + β)

2q∨qq

+ qqq) ,

x4 = α3

((1 + 2β)(1 + β)

6q∨qq q

+ (1 + β) q∨qqq+

(1 + β)2

q∨qq q + qqqq ) ,

x5 = α4

(1+3β)(1+2β)(1+β)

24q∨qq�H

q q + (1+2β)(1+β)2

q∨qq qq+ (1 + β)2 q∨qq∨qq

+(1 + β) q∨qqqq+ (1+2β)(1+β)

6q∨qq qq + (1+β)

2q∨qq qq

+ (1 + β) q∨qq qq + (1+β)2

qqq∨q q

+ qqqqq .

Proof. For any t ∈ T , we denote by bt the coefficient of t in xweight(t). Then b q = 1. Theformal series f(h) is given by:

f(h) =∞∑

n=0

αn [n]β !n!

hn.

If t = B+(ta11 . . . tak

k ), where t1, . . . , tk are distinct elements of T , then:

bt = αa1+...+ak[a1 + . . . + ak]β!(a1 + . . . + ak)!

(a1 + . . . + ak)!a1! . . . ak!

ba1t1

. . . baktk

.

The result comes from an easy induction. �

As a consequence, H0,β = K[ q ], so H0,β is a Hopf subalgebra. Moreover, Hα,β = H1,β ifα 6= 0: we can restrict ourselves to the case α = 1. In order to lighten the notations, we putnt = stet and [nt]β = [st]β[et]β for all t ∈ T . Then:{

n q = 1,nB+(t1...tk) = k!nt1 . . . ntk ,{

[n q ]β = 1,[nB+(t1...tk)]β = [k]β![nt1 ]β . . . [ntk ]β .

9

As a consequence:

nt =∏

s vertex of t

(fertility of s)!, [nt]β =∏

s vertex of t

[fertility of s]β !.

We shall use the following result, proved in [4, 6]:

Lemma 14 For all forests F ∈ F , G, H ∈ T , we denote by n(F,G;H) the coefficient ofF ⊗G in ∆(H), and n′(F,G;H) the number of graftings of the trees of F over G giving the treeH. Then n′(F,G;H)sH = n(F,G;H)sF sG.

Lemma 15 Let k, n ∈ N∗. We put, in K[X1, . . . , Xn], S = X1 + . . . + Xn. Then:∑α1+...+αn=k

n∏i=1

Xi(Xi + 1) . . . (Xi + αi − 1)αi

=S(S + 1) . . . (S + k − 1)

k!.

Proof. By induction on k, see [5]. �

Proposition 16 If α = 1:

∆(X) = X ⊗ 1 +∞∑

n=1

(1− βX)−n(1/β+1)+1 ⊗ xn.

So H1,β is a Hopf subalgebra.

Proof. As for all n ≥ 1, xn is a linear span of trees, we can write:

∆(X) = X ⊗ 1 +∑

F∈F , t∈TaF,tF ⊗ t.

Then, if F ∈ F , G ∈ T :

aF,G =∑H∈T

[nH ]βsH

n(F,G;H) =∑H∈T

[nH ]βsF sG

n′(F,G;H).

We put F = t1 . . . tk, and we denote by s1, . . . , sn the vertices of the tree G, of respective fertilityf1, . . . , fn. Let us consider a grafting of F over G, such that αi trees of F are grafted on thevertex si. Then α1 + . . . + αn = k. Denoting by H the result of this grafting:

[nH ]β = [nG]β [nt1 ]β . . . [ntk ]β[f1 + α1]β !

[f1]β !. . .

[fn + αn]β![fn]β !

.

Moreover, the number of such graftings isk!

α1! . . . αn!. So, with lemma 15, putting xi = fi + 1/β

and s = x1 + . . . + xn:

aF,G =∑

α1+...+αn=k

k!α1! . . . αk!

1sF sG

[nG]β[nt1 ]β . . . [ntk ]β[f1 + α1]β!

[f1]β!. . .

[fn + αn]β![fn]β!

=[nG]β !

sG

k![nt1 ]β . . . [ntk ]βsF

∑α1+...+αn=k

n∏i=1

(1 + fiβ) . . . (1 + (fi + αi − 1)β)αi!

=[nG]β !

sG

k![nt1 ]β . . . [ntk ]βsF

∑α1+...+αn=k

n∏i=1

βαixi(xi + 1) . . . (xi + αi + 1)

αi!

=[nG]β !

sG

k![nt1 ]β . . . [ntk ]βsF

βk∑

α1+...+αn=k

n∏i=1

xi(xi + 1) . . . (xi + αi + 1)αi!

=[nG]β !

sG

k![nt1 ]β . . . [ntk ]βsF

βk s(s + 1) . . . (s + k − 1)k!

.

10

Moreover, as G is a tree, s = f1 + . . . + fn + n/β = n− 1 + n/β = n(1 + 1/β)− 1.

We now write F = t1 . . . tk = ua11 . . . ual

l , where u1, . . . , ul are distinct elements of T . Then:

sF = sa1u1

. . . salul

a1! . . . al!,

so:k![nt1 ]β . . . [ntk ]β

sF=

(a1 + . . . + al)!a1! . . . al!

([nt1 ]βst1

)a1

. . .

([ntl ]βstl

)al

.

As a conclusion, putting Qk(S) =S(S + 1) . . . (S + k − 1)

k!:

∆(X) = X ⊗ 1 +∑n≥1

∑ta11 ...t

all ∈F

(a1 + . . . + al)!a1! . . . al!

βa1+...+alQa1+...+al(n(1 + 1/β)− 1)

([nt1 ]βst1

t1

)a1

. . .

([ntl ]βstl

tl

)al

⊗

∑G∈T

weight(G)=n

[nG]β!sG

G

= X ⊗ 1 +

∞∑n=1

(1− βX)−n(1/β+1)+1 ⊗ xn.

So ∆(X) ∈ H⊗H. Projecting on the homogeneous component of degree n, we obtain ∆(x) ∈H ⊗H, so H1,β is a Hopf subalgebra. �

Remarks.

1. For (α, β) = (1, 0), f(h) = eh and, for all n ∈ N∗, xn =∑t∈T

weight(t)=n

1st

t.

2. For (α, β) = (1, 1), f(h) = (1− h)−1 and, for all n ∈ N∗, xn =∑t∈T

weight(t)=n

ett.

3. For (α, β) = (1,−1), f(h) = 1+h and, as [i]−1 = 0 if i ≥ 2, for all n ∈ N∗, xn is the ladderof weight n.

2.4 What is Hα,β ?

If α = 0, then H0,β = K[ q ]. If α 6= 0, then obviously Hα,β = H1,β : let us suppose that α = 1.The Hopf algebra H1,β is graded, connected and commutative. Dually, its graded dual H∗1,β

is a graded, connected, cocommutative Hopf algebra. By the Milnor-Moore theorem [10], it isisomorphic to the enveloping algebra of the Lie algebra of its primitive elements. We now denotethis Lie algebra by g1,β . The dual of g1,β is identified with the quotient space:

coPrim(H1,β) =H1,β

(1)⊕Ker(ε)2,

and the transposition of the Lie bracket is the Lie cobracket δ induced by:

($ ⊗$) ◦ (∆−∆op),

where $ is the canonical projection on coPrim(H1,β). As H1,β is the polynomial algebra gen-erated by the xn’s, a basis of coPrim(H1,β) is ($(xn))n∈N∗ . By proposition 16:

($ ⊗$) ◦∆(X) = ($ ⊗$)

( ∞∑n=1

(1− βX)−n(1/β+1)+1 ⊗ xn

)=

∑n≥1

(n(1 + β)− β) $(X)⊗$(xn).

11

Projecting on the homogeneous component of degree k:

($ ⊗$) ◦∆(xk) =k∑

i+j=k

(j(1 + β)− β) $(xi)⊗$(xj).

As a consequence:δ($(xk)) =

∑i+j=k

(1 + β)(j − i)$(xi)⊗$(xj).

Dually, the Lie algebra g1,β has the dual basis (Zn)n≥1, with bracket given by:

[Zi, Zj ] = (1 + β)(j − i)Zi+j .

So, if β 6= −1, this Lie algebra is isomorphic to the Faa di Bruno Lie algebra gFdB, which has abasis (fn)n≥1, and its bracket defined by [fi, fj ] = (j− i)fi+j . So H1,β is isomorphic to the Hopfalgebra U(gFdB)∗, namely the Faa di Bruno Hopf algebra [3], coordinates ring of the group offormal diffeomorphisms of the line tangent to Id, that is to say:

GFdB =({∑

anhn ∈ K[[h]] / a0 = 0, a1 = 1}

, ◦)

.

Theorem 17 1. If α 6= 0 and β 6= −1, Hα,β is isomorphic to the Faa di Bruno Hopfalgebra.

2. If α 6= 0 and β = −1, Hα,β is isomorphic to the Hopf algebra of symmetric functions.

3. If α = 0, Hα,β = K[ q ].Remark. If β and β′ 6= −1, then H1,β and H1,β′ are isomorphic but are not equal, as it is

shown by considering x3.

3 FdB Lie algebras

In the preceding section, we considered Hopf subalgebra of H, generated in each degree by alinear span of trees. Their graded dual is then the enveloping algebra of a Lie algebra g, graded,with Poincare-Hilbert formal series:

h

1− h=

∞∑n=1

hn.

Under an hypothesis of non-commutativity, we show that such a g is isomorphic to the Faa diBruno Lie algebra, so the considered Hopf subalgebra is isomorphic to the Faa di Bruno Hopfalgebra.

Remark. The proves of these section were completed using MuPAD pro 4. The notebookof the computations can be found at http://loic.foissy.free.fr/pageperso/publications.html.

3.1 Definitions and first properties

Definition 18 Let g be a N-graded Lie algebra. For all n ∈ N, we denote by g(n) thehomogeneous component of degree n of g. We shall say that g is FdB if:

1. g is connected, that is to say g(0) = (0).

2. For all i ∈ N∗, g is one-dimensional.

3. For all n ≥ 2, [g(1), g(n)] 6= (0).

12

Let g be a FdB Lie algebra. For all i ∈ N∗, we fix a non-zero element Zi of g(i). By conditions1 and 2, (Zi)i≥1 is a basis of g. By homogeneity of the bracket of g, for all i, j ≥ 1, there existsan element λi,j ∈ K, such that:

[Zi, Zj ] = λi,jZi+j .

The Jacobi relation gives, for all i, j, k ≥ 1:

λi,jλi+j,k + λj,kλj+k,i + λk,iλk+i,j = 0. (3)

Moreover, by antisymmetry, λj,i = −λi,j for all i, j ≥ 1. Condition 2 is expressed as λ1,j 6= 0 forall j 6= 2.

Lemma 19 Up to a change of basis, we can suppose that λ1,j = 1 for all j ≥ 2 and thatλ2,3 ∈ {0, 1}.

Proof. We define a family of scalars by:α1 = 1,α2 6= 0,αn = λ1,2 . . . λ1,n−1α2 if n ≥ 3.

By condition 3, all these scalars are non-zero. We put Z ′i = αiZi. Then, for all j ≥ 2:

[Z ′1, Z′j ] = αjλ1,jZ1+j =

αjλ1,j

αj+1Z ′1+j = Z ′1+j .

So, replacing the Zi’s by the Z ′i’s, we can suppose that λ1,j = 1 if j ≥ 2.

Let us suppose now that λ2,3 6= 0. We then choose:

α2 =λ1,3λ1,4

λ2,3.

Then:

[Z ′2, Z′3] =

λ2,3α2α3

α5Z ′5 =

λ2,3α2λ1,2α2

λ1,2λ1,3λ1,4α2Z ′5 = Z ′5.

So, replacing the Zi’s by the Z ′i’s, we can suppose that λ2,3 = 1. �

Lemma 20 If i, j ≥ 2, λi,j =i−2∑k=0

(i− 2

k

)(−1)kλ2,j+k.

Proof. Let us write (3) with i = 1:

λ1,jλj+1,k + λj,kλj+k,1 + λk,1λk+1,j = 0.

If j, k ≥ 2, then λ1,j = −λk,1 = −λj+k,1 = 1, so:

λk+1,j = λk,j − λk,j+1. (4)

If k = 2, this gives the announced formula for i = 3.

13

Let us prove the result by induction on i. This is obvious for i = 2 and done for i = 3. Letus assume the result at rank i− 1. Then, by (4):

λi,j = λi−1,j − λi−1,j+1

=i−3∑k=0

(i− 3

k

)(−1)kλ2,j+k −

i−3∑k=0

(i− 3

k

)(−1)kλ2,j+1+k

=i−3∑k=0

(i− 3

k

)(−1)kλ2,j+k +

i−2∑k=1

(i− 3k − 1

)(−1)kλ2,j+k

= λ2,j +i−3∑k=1

(i− 2

k

)(−1)kλ2,j+k + (−1)i−2λ2,j+i−2

=i−2∑k=0

(i− 2

k

)(−1)kλ2,j+k.

So the result is true for all i ≥ 2. �

As a consequence, the λi,j ’s are entirely determined by the λ2,j ’s. We can ameliorate thisresult, using the following lemma:

Lemma 21 For all k ≥ 2, λ2,2k =1

2k − 3

2k−4∑l=0

(2k − 2

l

)(−1)lλ2,l+3.

Proof. Let us write the relation of lemma 20 for (i, j) = (3, 2k) and (i, j) = (2k, 3):

λ3,2k = λ2,2k − λ2,2k+1,

λ2k,3 =2k−2∑l=0

(2k − 2

l

)(−1)lλ2,3+l

= λ2,2k+1 − (2k − 2)λ2,2k +2k−4∑l=0

(2k − 2

l

)(−1)lλ2,3+l.

Summing these two relations:

−(2k − 3)λ2,2k +2k−4∑l=0

(2k − 2

l

)(−1)lλ2,3+l = 0.

This gives the announced result. �

As a consequence, the λi,j ’s are entirely determined by the λ2,j ’s, with j odd. In order tolighten the notations, we put µj = λ2,j for all j odd. Then, for example:

λ2,4 = µ3,λ2,6 = 2µ5 − µ3,λ2,8 = 3µ7 − 5µ5 + 3µ3,

λ2,10 = 4µ9 − 14µ7 + 28µ5 − 17µ3,λ2,12 = 5µ11 − 30µ9 + 126µ7 − 255µ5 + 155µ3.

Moreover, we showed that we can suppose that µ3 = 0 or 1.

14

Remark. The coefficient λ2,2k+4 is then a linear span of coefficients µ2i+3, 0 ≤ i ≤ k. Weput, for all k ∈ N:

λ2,2k+4 =k∑

i=0

ak,iµ2i+3.

We can prove inductively the following results:

1. For all k ∈ N, ak,k = k + 1.

2. For all k ≥ 1, ak,k−1 = −14

(2k+2

3

): up to the sign, this is the sequence A000330 of [11]

(pyramidal numbers).

3. For all k ≥ 2, ak,k−2 = 12

(2k+2

5

): this is the sequence A053132 of [11].

4. The sequence (−ak,0) is the sequence of signed Genocchi number, A001469 in [11].

It seems that for all i ≤ k:

ak,k−i =22i+2 − 1

i + 1B2i+2

(2k + 22i + 1

),

where the B2n’s are the Bernoulli number (see sequence A002105 of [11]).

3.2 Case where µ3 = 1

Lemma 22 Suppose that µ3 = 1. Then µ5 = 1 or910

.

Proof. By relation (3) for (i, j, k) = (2, 3, 4):

5µ5 − 3µ7 + µ5µ7 − 3 = 0.

If µ5 = 3, we obtain 12 = 0, absurd. So µ7 = −5µ5 − 3µ5 − 3

. By relation (3) for (i, j, k) = (2, 3, 6):

−2µ5 − 3

((2µ5 − 4)µ5µ9 + 3− 7µ5 + µ2

5 − 5µ35

)= 0.

If µ5 = 0, we obtain 2 = 0, absurd. If µ5 = 2, we obtain 66 = 0, absurd. So:

µ9 = −3− 7µ5 + µ25 − 5µ3

5

(2µ5 − 4)µ5.

Writing relation (3) for (i, j, k) = (3, 4, 5):

−9(µ5 − 1)5(10µ5 − 9)µ5(µ5 − 2)(µ5 − 3)2

= 0.

So µ5 = 1 or µ5 =910

. �

Proposition 23 Let us suppose that µ3 = µ5 = 1. Then:λ1,j = 1 if j ≥ 2,λ2,j = 1 if j ≥ 3,λi,j = 0 if i, j ≥ 3.

15

Proof. Let us first prove inductively on j that λ2,j = 1 if j ≥ 3. This is immediate if j = 3or 5 and comes from λ2,4 = µ3 for j = 4. Let us suppose the λ2,j = 1 for 3 ≤ j < n. If n = 2kis even, then:

λ2,2k =1

2k − 3

2k−4∑l=0

(2k − 2

l

)(−1)l = 1 +

12k − 3

2k−2∑l=0

(2k − 2

l

)(−1)l = 1.

If n = 2k + 1 is odd, write relation (3) for (i, j, k) = (2, 3, 2k − 2):

λ2,3λ5,2k−2 + λ3,2k−2λ2k+1,2 + λ2k−2,2λ2k,3 = 0,3∑

l=0

(3l

)(−1)lλ2,2k−2+l − λ2,2k−2 + λ2,2k−1 + λ2,2k−2(λ2,2k − λ2,2k+1) = 0,

λ2,2k−2 − 3λ2,2k−1 + 3λ2,2k − λ2,2k+1 + 1− λ2,2k+1 = 0,

1− 3 + 3− 2λ2,2k+1 + 1 = 0,

so λ2,2k+1 = 1. Finally, if i, j ≥ 3, λi,j =i−2∑k=0

(i− 2

k

)(−1)k = 0. �

Lemma 24 For all N ≥ 2, SN =N∑

l=0

(N

l

)(−1)l (l + 1)

(l + 2)(l + 3)=

N − 1(N + 3)(N + 2)(N + 1)

.

Proof. Indeed:

SN =N∑

l=0

N !(l + 3)!(N − l)!

(−1)l(l + 1)2

=1

(N + 3)(N + 2)(N + 1)

N+3∑j=3

(N + 3

j

)(−1)j(j − 2)2

=1

(N + 3)(N + 2)(N + 1)

N+3∑j=0

(N + 3

j

)(−1)j(j − 2)2

− 1(N + 3)(N + 2)(N + 1)

(4− (N + 3))

= 0 +N − 1

(N + 3)(N + 2)(N + 1).

�

Proposition 25 Let us suppose that µ3 = 1 and µ5 =910

. Then, for all i, j ≥ 1:

λi,j =6(i− j)(i− 2)!(j − 2)!

(i + j − 2)!.

Proof. We first prove that λ2,n =6(n− 2)(n− 1)n

. This is immediate for n = 1, 2, 3, 4, 5. Let us

suppose the result for all j < n, with n ≥ 6. If n = 2k is even, using lemma 20:

λ2,2k =6

2k − 3

2k−4∑l=0

(2k − 2

l

)(−1)l l + 1

(l + 2)(l + 3).

Then lemma 24 gives the result. If n = 2k + 3 is odd, let us write the relation (3) with(i, j, k) = (2, 3, 2k):

λ2,3λ5,2k + λ3,2kλ2k+3,2 + λ2k,2λ2k+2,3 = 0.

16

So, with relation (4):

λ2,2k − 3λ2,2k+1 + 3λ2,2k+2 − λ2,2k+3 − λ2,2k+3(λ2,2k − λ2,2k+1) + λ2,2k(λ2,2k+2 − λ2,2k+3) = 0,

λ2,2k+3(−1− 2λ2,2k + λ2,2k+1) + λ2,2k − 3λ2,2k+1 + 3λ2,2k+2 + λ2,2kλ2,2k+2 = 0,

−λ2,2k+3(2k + 3)(k − 1)(2k + 5)

k(2k + 1)(2k − 1)+

3(2k + 5)(k − 1)k(k + 1)(2k − 1)

= 0,

which implies the result.Let us now prove the result by induction on i. This is immediate if i = 1, and the first part

of this proof for i = 2. Let us suppose the result at rank i. Then, by relation (4):

λi+1,j = λi,j − λi,j+1

= 6(j − i)(i− 2)!(j − 2)!

(i + j − 2)!− 6

(j + 1− i)(i− 2)!(j − 1)!(i + j − 1)!

= 6(i− 1)!(j − 2)!(j + 1− i)

(i + j − 1)!.

So the result is true for all i, j. �

3.3 Case where µ3 = 0

Proposition 26 If µ3 = 0, then λi,j = 0 for all i, j ≥ 2.

Proof. We first prove that µ5 = 0. If not, by (3) for (i, j, k) = (2, 3, 4), µ5µ7 = 0, soµ7 = 0. By (3) with (i, j, k) = (2, 3, 7), −5µ5(28µ5 + 4µ9) = 0, so µ9 = −7µ5. By (3) with(i, j, k) = (3, 4, 5), −36µ2

5 = 0: contradiction. So µ5 = 0.Let us then prove that all the µ2k+1’s, k ≥ 1, are zero. We assume that µ3 = µ5 = . . . =

µ2k−1 = 0, and µ2k+1 6= 0, with l ≥ 3. By lemma 21, λ2,2 = . . . = λ2,2k−1 = λ2,2k = 0 andλ2,2k+1 6= 0. By relation (3) for (i, j, k) = (2, 3, n), combined with (4):

λ2,3λ5,n + λ3,nλn+3,2 + λn,2λn+2,3 = 0,

−(λ2,n − λ2,n+1)λ2,n+3 + λ2,n(λ2,n+2 − λ2,n+3) = 0.

For n = 2k, this gives λ2,2k+1λ2,2k+3 = 0, so λ2,2k+3 = 0. For n = 2k + 2:

λ2,2k+2(λ2,2k+4 − 2λ2,2k+5) = 0. (5)

By lemma 21:

λ2,2k+2 =1

2k − 1

(2k

2k − 2

)λ2,2k+1

= kλ2,2k+1,

λ2,2k+4 =1

2k + 1

((2k + 2

2k

)λ2,2k+3 −

(2k + 22k − 1

)λ2,2k+2 +

(2k + 22k − 2

)λ2,2k+1

)= −k(k + 1)(2k + 1)

6λ2,2k+1.

With (5):

λ2,2k+5 = −k(k + 1)(2k + 1)12

λ2,2k+1.

By relation (3) for (i, j, k) = (3, 4, 2k):

λ3,4λ7,2k + λ4,2kλ4+2k,3 + λ2k,3λ2k+3,4 = 0.

17

Moreover, using lemma 20:λ3,4 = λ2,4 − λ2,5 = 0,

λ4,2k = λ2,2k − 2λ2,2k+1 + λ2,2k+2,λ3,4+2k = λ2,4+2k − λ2,5+2k,

λ2k,3 = −λ2,2k + λ2,2k+1,λ3+2k,4 = −λ2,3+2k + 2λ2,4+2k − λ2,5+2k.

This gives:λ2

2,2k+1k(3k − 11)(2k + 1)(k + 1)12

= 0,

so λ2,2k+1 = µ2k+1 = 0: contradiction. So all the µ2k+1, k ≥ 1, are zero. By lemma 21, theλ2,i’s, i ≥ 2, are zero. By lemma 20, the λi,j ’s, i, j ≥ 2, are zero. �

Theorem 27 Up to an isomorphism, there are three FdB Lie algebras:

1. The Faa di Bruno Lie algebra gFdB, with basis (ei)i≥1, and bracket given by [ei, ej ] =(j − i)ei+j for all i, j ≥ 1.

2. The corolla Lie algebra gc, with basis (ei)i≥1, and bracket given by [e1, ej ] = ej+1 and[ei, ej ] = 0 for all i, j ≥ 2.

3. Another Lie algebra g3, with basis (ei)i≥1, and bracket given by [e1, ei] = ei+1, [e2, ej ] =ej+2, and [ei, ej ] = 0 for all i ≥ 2, j ≥ 3.

Proof. We have first to prove that these are indeed Lie algebras: this is done by directcomputations. Let g be a FdB Lie algebra. We showed that three cases are possible:

1. µ3 = 1 and µ5 =910

. By proposition 25, putting ei =Zi

6(i− 2)!if i ≥ 2 and e1 = Z1, we

obtain the Faa di Bruno Lie algebra.

2. µ3 = µ5 = 1. By proposition 23, we obtain the third Lie algebra.

3. µ3 = 0. By proposition 26, we obtain the corolla Lie algebra.

�

Corollary 28 Let g be a FdB Lie algebra, such that if i and j are two distinct elements ofN∗, then [g(i), g(j)] 6= (0). Then g is isomorphic to the Faa di Bruno Lie algebra.

4 Dual of enveloping algebras of FdB Lie algebras

We realized in the first section the Faa di Bruno Hopf algebra, dual of the enveloping algebra ofthe Faa di Bruno Lie algebra, as a Hopf subalgebra of H. We now give a similar result for thetwo other FdB Lie algebra.

4.1 The corolla Lie algebra

Definition 29 We denote by Hc the subalgebra of H generated by the corollas.

Proposition 30 Hc is a graded Hopf subalgebra of H. Its dual is isomorphic to the envelop-ing algebra of the corolla Lie algebra.

18

Proof. The subalgebra Hc, being generated by homogeneous elements, is graded. By lemma9, Hc is a Hopf subalgebra of H. As it is commutative, its dual is the enveloping algebra ofthe Lie algebra Prim(H∗c). The dual of this Lie algebra is the Lie coalgebra coPrim(Hc) =

Hc

(1)⊕Ker(ε)2, with cobracket δ induced by ($ ⊗ $) ◦ (∆ − ∆op). As Hc is generated by the

corollas, a basis of coPrim(Hc) is ($(cn))n≥1. Moreover, if n ≥ 1:

($ ⊗$) ◦∆(cn) = $(c1)⊗$(cn−1),δ(cn) = $(c1)⊗$(cn−1)−$(cn−1)⊗$(c1).

Let (Zn)n≥1 be the basis of Prim(H∗c), dual of the basis ($(cn))n≥1. By duality, for all i, j ∈ N∗,such that i 6= j:

[Zi, Zj ] =

Z1+j if i = 1,−Zi+1 if j = 1,0 otherwise.

So Prim(H∗c) is isomorphic to the corolla Lie algebra, via the morphism:{gc −→ Prim(H∗c)ei −→ Zi.

Dually, Hc is isomorphic to U(gc)∗. �

Remark. We work in K[T ][β]. The generators of H1,β then satisfy:

xn+1 =[n]β !n!

cn+1 +O(βn−2).

Note that the degree of [n]β! in β is n− 1. So:

limβ→∞

n![n]β!

xn+1 = cn+1.

In this sense, the Hopf algebra Hc is the limit of H1,β when β goes to infinity.

4.2 The third FdB Lie algebra

We consider the following element of H:

Y = B+

(exp

( qq − 12

q2 + q)) =∑n≥1

yn.

For example:

y1 = qy2 = qq ,y3 = qqq ,y4 = q∨qqq

− 13

q∨qq q,

y5 =12

q∨qq qq− 1

12q∨qq�Hq q

.

Definition 31 We denote by H3 the subalgebra of H generated by the yn’s.

Proposition 32 H3 is a graded Hopf subalgebra of H. Its dual is isomorphic to the en-veloping algebra of the third FdB Lie algebra.

19

Proof. The subalgebra H3, being generated by homogeneous elements, is graded. An easy

computation proves that X = qq − 12

q2 + q is a primitive element of H. As a consequence, in H,

by (1):

∆(X) = X ⊗ 1 + 1⊗X,

∆(exp(X)) = exp(X ⊗ 1 + 1⊗X)= exp(X ⊗ 1) exp(1⊗X)= (exp(X)⊗ 1)(1⊗ exp(X))= exp(X)⊗ exp(X),

∆(Y ) = ∆ ◦B+(exp(X))= Y ⊗ 1 + exp(X)⊗ Y.

Moreover, X = y2−12y21 +y1 ∈ H3, so taking the homogeneous component of degree n of ∆(Y ),

we obtain:

∆(yn) = yn ⊗ 1 +n∑

k=1

n−k∑l=1

∑a1+...+al=n−k

1l!

xa1 . . . xal⊗ yk,

where x1 = q = y1, x2 = qq − 12

q q = y2− 12y2

1 and xi = 0 if i ≥ 3, so ∆(yn) ∈ H3⊗H3 and H3 is aHopf subalgebra of H. As it is commutative, its dual is the enveloping algebra of the Lie algebra

Prim(H∗3). The dual of this Lie algebra is the Lie coalgebra coPrim(H3) =H3

(1)⊕Ker(ε)2,

with cobracket δ induced by ($ ⊗ $) ◦ (∆ −∆op). As H3 is generated by the yn’s, a basis ofcoPrim(H3) is ($(yn))n≥1. Moreover:

($ ⊗$) ◦∆(Y ) = $(exp(X))⊗$(Y )= $(X)⊗$(Y )= ($(y2) + $(y1))⊗$(Y ).

Taking the homogeneous component of degree n, with the convention y−1 = y0 = 0:

($ ⊗$) ◦∆(yn) = $(y2)⊗$(yn−2) + $(y1)⊗$(yn−1),δ($(yn)) = $(y2)⊗$(yn−2) + $(y1)⊗$(yn−1)

−$(yn−2)⊗$(y2)−$(yn−1)⊗$(y1).

Let (Zn)n≥1 be the basis of Prim(H∗3), dual of the basis ($(cn))n≥1. By duality, for all i, j ∈ N∗,such that i ≥ 2 and j ≥ 3:

[Z1, Zi] = Z1+j ,[Z2, Zj ] = Z2+j ,[Zi, Zj ] = 0.

So Prim(H∗3) is isomorphic to third FdB Lie algebra, via the morphism:{g3 −→ Prim(H∗c)ei −→ Zi.

Dually, H3 is isomorphic to U(g3)∗. �

References

[1] Christoph Bergbauer and Dirk Kreimer, Hopf algebras in renormalization theory: localityand Dyson-Schwinger equations from Hochschild cohomology, IRMA Lect. Math. Theor.Phys., vol. 10, Eur. Math. Soc., Zrich, 2006, arXiv:hep-th/05 06190.

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[2] Alain Connes and Dirk Kreimer, Hopf algebras, Renormalization and Noncommutative ge-ometry, Comm. Math. Phys 199 (1998), no. 1, 203–242, arXiv:hep-th/98 08042.

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[8] , Dyson-Schwinger equations: from Hopf algebras to number theory, Universality andrenormalization, Fields Inst. Commun., no. 50, Amer. Math. Soc., Providence, RI, 2007,arXiv:hep-th/06 09004.

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[10] John W. Milnor and John C. Moore, On the structure of Hopf algebras, Ann. of Math. (2)81 (1965), 211–264.

[11] N. J. A Sloane, On-line encyclopedia of integer sequences,http://www.research.att.com/ njas/sequences/Seis.html.

[12] Richard P. Stanley, Enumerative combinatorics. Vol. 1., Cambridge Studies in AdvancedMathematics, no. 49, Cambridge University Press, Cambridge, 1997.

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